I'm trying to solve the very same problem stated in this question: Conditional independence in Bayesian network with qualitative influences
Take for example pair 5: P(c=1|h=0) vs P(c=1)
I tried to solve this by using the formula
$p(\textbf{x}) = \prod p(x|parentof(x))$
Which resulted in:
$P(c^{1}) = P(c^{1}|d,h) = P(c^{1}|d^{1},h^{0}) + P(c^{1}|d^{0},h^{0}) + P(c^{1}|d^{1},h^{1}) + P(c^{1}|d^{0},h^{1}) $
and
$P(c^{1}|h^{0}) = \frac{P(c^{1},h^{0})}{P(h^{0})} = \frac{P(h^{0})P(c^{1}|d,h^{0})}{P(h^{0})} = P(c^{1}|d^{1},h^{0}) + P(c^{1}|d^{0},h^{0}) $
where I used the above mentioned formula in the second step.
In the alternative version of this problem I'm working on we are assuming that no probabilities are 0 or 1. My conclusion was thus that $P(c^{1}) > P(c^{1}|h^{0}) $ since $P(c^{1})$ contains the same terms as $P(c^{1}|h^{0})$ and two additional ones. I find this to be counterintuitive since observing a negative result of health consciousness would act in favor of observing a negative result on good diet and thus make it more probable that we have high cholesterol. I assume I have made some errors in my calculations and would be very happy if someone could point them out!
You have the formula wrong. The Law of Total Probability is:
$$\begin{align}\mathsf p(\mathbf x) & = \prod \mathsf p(\mathbf x, \operatorname{parentof}(\mathbf x)) \\[1ex] & =\prod \mathsf p(\mathbf x\mid \operatorname{parentof}(\mathbf x))\cdot\mathsf p(\operatorname{parentof}(\mathbf x))\end{align}$$
So $$\begin{align}\mathsf P(c^1) & = \prod_{x,y} \mathsf P(c^1\mid d^x,h^y) \\[1ex] & ={\mathsf P(c^1\mid d^0,h^0)~\mathsf P(d^0,h^0)+\mathsf P(c^1\mid d^0,h^1)~\mathsf P(d^0,h^1)+\mathsf P(c^1\mid d^1,h^0)~\mathsf P(d^1,h^0)+\mathsf P(c^1\mid d^1,h^1)~\mathsf P(d^1,h^1)}\end{align}$$